Proofs of Trigonometric Identities

Here are some tips to prove trigonometric identities.

• Work only on one side of the equation.

• Convert everything in terms of sine and/or cosine. This simplifies the expression.

• Try to obtain one or more of the basic trigonometric identities.

• Put any fractions over a common denominator.

• Remove the greatest common factor, or use other types of factoring to simplify the terms.

• Multiply the numerator and denominator by the conjugate of either the numerator or denominator.

Prove the following identity: |\sin^2\theta=1-\cot^2\theta\,\sin^2\theta.|

Simplify the right-hand side.

Start by rewriting everything in terms of |\sin| and |\cos,| and then simplify the resulting expression.||\begin{align}\sin^2\theta&=1-\cot^2\theta\,\sin^2\theta\\\\\sin^2\theta&=1-\dfrac{\cos^2\theta}{\color{#ec0000}{\cancel{\color{black}{\sin^2\theta}}}}\color{#ec0000}{\cancel{\color{black}{\sin^2\theta}}}\\\\\sin^2\theta&=1-\cos^2\theta\end{align}||By isolating |\sin^2\theta| in the 1st Pythagorean identity, we get the equation |\sin^2\theta=1-\cos^2\theta.| Substitute |1-\cos^2\theta| with |\sin^2\theta| on the right-hand side of the equation.||\sin^2\theta=\sin^2\theta||Both sides of the equation are identical. This proves the identity |\sin^2\theta=1-\cot^2\theta\,\sin^2\theta.|

Prove the following identity: |\dfrac{\sec x}{\cot x}=\dfrac{\sin x}{1-\sin^2 x}.|

This time, let’s work with the left-hand side of the equation. Rewrite it in terms of |\sin| and |\cos.|||\begin{align}\dfrac{\sec x}{\cot x}&=\dfrac{\sin x}{1-\sin^2 x}\\\\\dfrac{\dfrac{1}{\cos x}}{\dfrac{\cos x}{\sin x}}&=\dfrac{\sin x}{1-\sin^2 x}\end{align}||Divide the 2 fractions. Dividing by a fraction is equivalent to multiplying by its reciprocal.||\begin{align}\dfrac{1}{\cos x}\times\dfrac{\sin x}{\cos x}&=\dfrac{\sin x}{1-\sin^2 x}\\\\\dfrac{\sin x}{\cos^2 x}&=\dfrac{\sin x}{1-\sin^2 x}\end{align}||Isolating |\cos^2 x| from the identity |\cos^2 x+\sin^2 x=1,| gives the equation |\cos^2 x=1-\sin^2 x.| Substitute |\cos^2 x| with |1-\sin^2 x| in the fraction.||\begin{align}\dfrac{\sin x}{\boldsymbol{\color{#3a9a38}{\cos^2 x}}}=\dfrac{\sin x}{1-\sin^2 x}\\\\\dfrac{\sin x}{\boldsymbol{\color{#3a9a38}{1-\sin^2 x}}}=\dfrac{\sin x}{1-\sin^2 x}\end{align}||Both sides of the equation are identical. This proves the identity |\dfrac{\sec x}{\cot x}=\dfrac{\sin x}{1-\sin^2 x}.|

There is often more than one way to prove a trigonometric identity. In the previous example, we could have also worked with the right-hand side of the equation.||\begin{align}\dfrac{\sec x}{\cot x}&=\dfrac{\sin x}{1-\sin^2 x}\\\\\dfrac{\sec x}{\cot x}&=\dfrac{\sin x}{\cos^2 x}\\\\\dfrac{\sec x}{\cot x}&=\dfrac{1}{\cos x}\times\dfrac{\sin x}{\cos x}\\\\\dfrac{\sec x}{\cot x}&=\sec x\times\dfrac{1}{\cot x}\\\\\dfrac{\sec x}{\cot x}&=\dfrac{\sec x}{\cot x}\end{align}||

Prove the following identity: |\cos\theta\,\sqrt{\sec^2\theta-1}=\sin\theta.|

Let’s work with the left-hand side of the equation. Using the identity |1+\tan^2\theta=\sec^2\theta,| we get the equation |\tan^2\theta=\sec^2\theta-1.| We then substitute this in the left-hand side of the equation and simplify.||\begin{align}\cos\theta\,\sqrt{\boldsymbol{\color{#3a9a38}{\sec^2\theta-1}}}&=\sin\theta\\\cos\theta\,\sqrt{\boldsymbol{\color{#3a9a38}{\tan^2\theta}}}&=\sin\theta\\\cos\theta\,\tan\theta&=\sin\theta\end{align}||Replace tangent with the ratio |\dfrac{\sin\theta}{\cos\theta}| and simplify the expression obtained.||\begin{align}\cancel{\cos\theta}\,\dfrac{\sin\theta}{\cancel{\cos\theta}}&=\sin\theta\\\sin\theta&=\sin\theta\end{align}||Both sides of the equation are identical. This proves the identity |\cos\theta\,\sqrt{\sec^2\theta-1}=\sin\theta.|

Prove the following identity: |\dfrac{\cos^2\theta\,\tan\theta}{\cot\theta}=\sin^2\theta.|

Let’s work with the left-hand side of the equation. Rewrite it in terms of |\sin| and |\cos,| then simplify the resulting expression.||\begin{align}\dfrac{\cos^2\theta\,\tan\theta}{\cot\theta}&=\sin^2\theta\\\\\dfrac{\cos^2\theta\times\dfrac{\sin\theta}{\cos\theta}}{\dfrac{\cos\theta}{\sin\theta}}&=\sin^2\theta\\\\\cancel{\cos^2\theta}\times\dfrac{\sin\theta}{\cancel{\cos\theta}}\times\dfrac{\sin\theta}{\cancel{\cos\theta}}&=\sin^2\theta\\\\\sin^2\theta&=\sin^2\theta\end{align}||Both sides of the equation are identical. This proves the identity |\dfrac{\cos^2\theta\,\tan\theta}{\cot\theta}=\sin^2\theta.|

Prove the following identity: |\tan^2\theta-\sin^2\theta=\tan^2\theta\,\sin^2\theta.|

Work with the left-hand side of the equation. Rewrite it in terms of |\sin| and |\cos,| then simplify the resulting expression.

||\begin{align}\tan^2\theta-\sin^2\theta&=\tan^2\theta\,\sin^2\theta\\\\\dfrac{\sin^2\theta}{\cos^2\theta}-\sin^2\theta&=\tan^2\theta\,\sin^2\theta\\\\\dfrac{\sin^2\theta}{\cos^2\theta}-\dfrac{\sin^2\theta\,\cos^2\theta}{\cos^2\theta}&=\tan^2\theta\,\sin^2\theta\\\\\dfrac{\sin^2\theta-\sin^2\theta\,\cos^2\theta}{\cos^2\theta}&=\tan^2\theta\,\sin^2\theta\end{align}||

Factor out |\sin^2\theta| from the numerator.||\dfrac{\sin^2\theta\,(1-\cos^2\theta)}{\cos^2\theta}=\tan^2\theta\,\sin^2\theta||Use the fact that |\cos^2\theta+\sin^2\theta=1| to replace |1-\cos^2\theta| with |\sin^2\theta.| Next, group the factors in a way that allows |\dfrac{\sin^2\theta}{\cos^2\theta}| to be replaced with |\tan^2\theta.|||\begin{align}\dfrac{\sin^2\theta\,(\boldsymbol{\color{#3a9a38}{1-\cos^2\theta}})}{\cos^2\theta}&=\tan^2\theta\,\sin^2\theta\\\\\dfrac{\sin^2\theta\,\boldsymbol{\color{#3a9a38}{\sin^2\theta}}}{\cos^2\theta}&=\tan^2\theta\,\sin^2\theta\\\\\boldsymbol{\color{#3a9a38}{\dfrac{\sin^2\theta}{\cos^2\theta}}}\,\sin^2\theta&=\tan^2\theta\,\sin^2\theta\\\\\boldsymbol{\color{#3a9a38}{\tan^2\theta}}\,\sin^2\theta&=\tan^2\theta\,\sin^2\theta\end{align}||Both sides of the equation are identical. We have successfully proven the identity |\tan^2\theta-\sin^2\theta=\tan^2\theta\,\sin^2\theta.|

Prove the following identity: |\sin\theta\,(1+\tan\theta)+\cos\theta\,(1+\cot\theta)=\csc+\sec\theta.|

Rewrite the entire left-hand side of the equation in terms of |\sin| and |\cos,| then multiply.||\begin{align}\sin\theta\,(1+\tan\theta)+\cos\theta\,(1+\cot\theta)&=\csc\theta+\sec\theta\\\\\sin\theta\left(1+\dfrac{\sin\theta}{\cos\theta}\right)+\cos\theta\left(1+\dfrac{\cos\theta}{\sin\theta}\right)&=\csc\theta+\sec\theta\\\\\sin\theta+\dfrac{\sin^2\theta}{\cos\theta}+\cos\theta+\dfrac{\cos^2\theta}{\sin\theta}&=\csc\theta+\sec\theta\end{align}||Rearrange the terms, put the first 2 over a common denominator and the last 2 over another common denominator.||\begin{align}\sin\theta+\dfrac{\cos^2\theta}{\sin\theta}+\cos\theta+\dfrac{\sin^2\theta}{\cos\theta}&=\csc\theta+\sec\theta\\\\\dfrac{\sin^2\theta}{\sin\theta}+\dfrac{\cos^2\theta}{\sin\theta}+\dfrac{\cos^2\theta}{\cos\theta}+\dfrac{\sin^2\theta}{\cos\theta}&=\csc\theta+\sec\theta\\\\\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta}+\dfrac{\cos^2\theta+\sin^2\theta}{\cos\theta}&=\csc\theta+\sec\theta\end{align}||Use the identity |\cos^2\theta+\sin^2\theta=1| to simplify the numerators.||\begin{align}\dfrac{1}{\sin\theta}+\dfrac{1}{\cos\theta}&=\csc\theta+\sec\theta\\\\\csc\theta+\sec\theta&=\csc\theta+\sec\theta\end{align}||Both sides of the equation are identical. This proves the identity |sin\theta\,(1+\tan\theta)+\cos\theta\,(1+\cot\theta)=\csc+\sec\theta.|

Prove the following identity: |(\csc\,\theta-\cot\theta)^2=\dfrac{1-\cos\theta}{1+\cos\theta}.|

Expand the square on the left-hand side of the equation.||\begin{align}(\csc\theta-\cot\theta)^2&=\dfrac{1-\cos\theta}{1+\cos\theta}\\\\(\csc\theta-\cot\theta)(\csc\theta-\cot\theta)&=\dfrac{1-\cos\theta}{1+\cos\theta}\\\\\csc^2\theta-2\,\csc\theta\,\cot\theta+\cot^2\theta&=\dfrac{1-\cos\theta}{1+\cos\theta}\end{align}||Rewrite it in terms of |\sin| and |\cos.|||\begin{align}\csc^2\theta-2\csc\theta\,\cot\theta+\cot^2\theta&=\dfrac{1-\cos\theta}{1+\cos\theta}\\\\\dfrac{1}{\sin^2\theta}-2\times\dfrac{1}{\sin\theta}\times\dfrac{\cos\theta}{\sin\theta}+\dfrac{\cos^2\theta}{\sin^2\theta}&=\dfrac{1-\cos\theta}{1+\cos\theta}\\\\\dfrac{1}{\sin^2\theta}-\dfrac{2\cos\theta}{\sin^2\theta}+\dfrac{\cos^2\theta}{\sin^2\theta}&=\dfrac{1-\cos\theta}{1+\cos\theta}\end{align}||Since the terms all have a common denominator, group them and rearrange them in descending order of degree.||\dfrac{\cos^2\theta-2\cos\theta+1}{\sin^2\theta}=\dfrac{1-\cos\theta}{1+\cos\theta}||Using the identity |\cos^2\theta+\sin^2\theta=1,| gives the equation |\sin^2\theta=1-\cos^2\theta.| So, replace |\sin^2\theta.|||\dfrac{\cos^2\theta-2\cos\theta+1}{\boldsymbol{\color{#3a9a38}{1-\cos^2\theta}}}=\dfrac{1-\cos\theta}{1+\cos\theta}||Now factor both the numerator and the denominator. The numerator is a perfect square trinomial, while the denominator is a difference of squares.

||\dfrac{(\cos\theta-1)(\cos\theta-1)}{(1-\cos\theta)(1+\cos\theta)}=\dfrac{1-\cos\theta}{1+\cos\theta}||To simplify the binomials, factor out |-1| from both pairs of brackets in the numerator.||\begin{align}\dfrac{(-1)(-\cos\theta+1)\,(-1)(-\cos\theta+1)}{(1-\cos\theta)(1+\cos\theta)}&=\dfrac{1-\cos\theta}{1+\cos\theta}\\\dfrac{\overbrace{(-1)(-1)}^{\large{=1}}(1-\cos\theta)(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)}&=\dfrac{1-\cos\theta}{1+\cos\theta}\\\\\dfrac{\cancel{(1-\cos\theta)}(1-\cos\theta)}{\cancel{(1\cos\theta)}(1+\cos\theta)}&=\dfrac{1-\cos\theta}{1+\cos\theta}\\\\\dfrac{1-\cos\theta}{1+\cos\theta}&=\dfrac{1-\cos\theta}{1+\cos\theta}\end{align}||Both sides of the equation are identical. This proves the identity |(\csc\theta-\cot\theta)^2=\dfrac{1-\cos\theta}{1+\cos\theta}.|